Math 324 - Section 4.4 - -The Exponential Distribution with parameter-The PDF is = with support \u2265 0 The CDF is \u2264 = 11 \u2212 \u2265 0 0 < 0 For

Math 324 - Section 4.4 - -The Exponential Distribution...

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--------The Exponential Distribution with parameter ࠵? ----- The PDF is ࠵?(࠵?) = ࠵?࠵? )*+ with support ࠵? ≥ 0 . The CDF is ࠵?(࠵? ≤ ࠵?) = 1 1 − ࠵? )*+ ࠵? ≥ 0 0 ࠵? < 0 Integration by parts shows that ࠵? = ࠵?(࠵?) = 8 ࠵?࠵?࠵? )*+ ࠵?࠵? = 1 ࠵? : ; ࠵? = = ࠵?(࠵? = ) − ࠵? = = 8 ࠵? = ࠵?࠵? )*+ ࠵?࠵? − ࠵? )= = 1 ࠵? = : ; Notice that the CDF is completely determined by the so called “tail distribution function” or TDF ࠵?(࠵? > ࠵?) = ࠵? )*+ for ࠵? ≥ 0 Proposition: For a time homogeneous Poisson process with rate parameter ࠵? per unit time, the waiting time ࠵? @ until the first occurrence has exponential distribution with parameter ࠵? . Proof: For any nonnegative ࠵? the TDF is ࠵?(࠵? @ > ࠵?) = ࠵?(there are zero occurrences in interval of length ࠵? ) = ࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?࠵?(࠵?࠵?, 0) = ࠵? )*+ Therefore the CDF ࠵?(࠵? @ ≤ ࠵?) matches with the CDF of the exponential distribution. Proposition (Memoryless property ): For exponential distribution ࠵?(࠵? > ࠵? + ࠵? ; | ࠵? > ࠵? ; ) = ࠵?(࠵? > ࠵?) (assuming ࠵?, ࠵? ; > 0) Proof: ࠵?(࠵? > ࠵? + ࠵? ; | ࠵? > ࠵? ; ) = ࠵?({࠵? > ࠵? + ࠵? ; } ∩ {࠵? > ࠵? ; }) ࠵?(࠵? > ࠵? ; ) = ࠵?(࠵? > ࠵? + ࠵? ; ) ࠵?(࠵? > ࠵? ; ) = ࠵? )*(+\+ ] ) ࠵? )*+ ] = ࠵? )*+ Example: Let X have exponential distribution with parameter ࠵? = 1.6 . 1. The 70 th percentile solves ࠵?(࠵? ≤ ࠵?) = 0.70, so that ࠵?(࠵? > ࠵?) = ࠵? )@.a+ = 0.30 . Solving this equation with natural log gives ࠵? = 0.75248 For exponential distributions: The mean is the reciprocal of the parameter. The variance is the square of the mean. The SD is equal to the mean.

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